##### VMC VIQ Scholarship Test

ApplyRegister for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 02:35 PM IST

By regularly practicing RD Sharma Class 12th Exercise 17.3, understudies will be severe with the thoughts given in the Class 12 RD Sharma textbook. Be that as it may, it is nearly impossible to be skilled at maths who thought proper practice. The Rd Sharma class 12th exercise 17.3 will help students to learn and practice at home so that they have the chance to improve their skills to score well in boards and JEE mains.

**JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days**

**JEE Main 2025: Maths Formulas | Study Materials**

**JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics **

This Story also Contains

- RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise
- Maxima and Minima Excercise: 17.3
- Maxima and Minima excercise 17.3 question 1 (iii)
- Maxima and Minima Excercise 17.3 question 1 (v)
- Maxima and Minima exercise 17.3 question 1 (vi)
- Maxima and Minima excercise 17.3 question 1 (ix)Answer:
- Maxima and Minima exercise 17.3 question 2 (i)
- Maxima and Minima exercise 17.3 question 3
- Maxima and Minima excercise 17.3 question 4
- Maxima and Minima exercise 17.3 question 8
- Benefits of picking RD Sharma Arithmetic Solutions from Career360 include:
- RD Sharma Chapter-wise Solutions

Maxima and Minima excercise 17.3 question 1 (i)

Point of local maxima is at x=1 and it’s local maximum value is 68. Also point of local minima is at x=5,-6 and it’s local minimum values are -316 and -1647.

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value

It is given that

To find critical values ,

On solving,

critical values of x are 5,1,-6.

Now,

At x=5,

So, x=5 is point of local minima.

At x=1,

So, x=1 is point of local maxima.

At x=-6,

So, x=-6 is a point of local minima.

The local max. value at x=1 is

And local min. value at x=5 & x=-6 are,

Hence, point of local maxima is 1 & its maximum value is 68.

Also, point of local minima are 5&-6 and its minimum values are -316 &-1647 respectively.

Maxima and Minima excercise 17.3 question 1 (ii)**Answer:**

point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.**Hint:**

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.**Given:****Explanation:**

We have,

To find maxima and minima.

At x = 3,

f’’(3) = 6(3)-12

=6>0

So, x = 3 is point of local minima

At x = 1,

f’’(1) = 6(1)-12

=-6<0

So, x = 1 is point of local maxima

Now, local maximum value at x= 1 is

And local min. value at x=3 is

Thus, point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.

Point of local maxima value is -2 and it’s local maximum value is 0. Also point of local minima is 0and it’s local minimum value is -4

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.

We have,

For max and min, f’(x)=0,

At x=0,

So, x= 0 is point of local minima

At x=-2,

So, x = -2 is point of local maxima

So, local max. value at x=-2 is

f(-2)=(-2-1)(-2+2)^{2}=0

And local min. value at x= 0 is

Thus, point of local maxima is -2 & its max. value is 0 & point of local minima is 0 & its value is -4.

Maxima and Minima exercise 17.3 question 1 (iv)

**Answer:**

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.

Given:

Explanation:

We have,

For maxima and minima, f’(x)=0

At x=2,

So, x=2 is point of local maxima.

So, local max. value at x= 2 is

Hence, point of local maxima is 2 & its max. value is

**Answer:**

Point of local minima value is x=-1 and it’s local minimum value is **Hint:**

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.**Given**:**Explanation:**

We have,

Now, for maxima & minima, f’(x)=0

at x=-1,

X = -1 is point of local minima.

So, local min. value at x= 1 is

Thus, point of local minima is -1 & its min value is

Point of local minima value is 2 and it’s local minimum value is 2.

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.

We have,

For local maxima or minima, we have

since,

Now, at x=2 ,

Thus, its min. value at 2 is 2.

Maxima and Minima Excercise 17.3 question 1 (vii)**Answer:**

Point of local minima value is and it’s local minimum value is **Hint:**

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.**Given:****Explanation: **

we have ,

For maxima & minima, f’(x)=0

At x=

So , is a point of local minima

Now at

Thus, point of local minima is & its value is

Maxima and Minima exercise 17.3 question 1 (viii)

Point of local maxima value is 4 and it’s local maximum value is 16. Also point of local minima is -4 and it’s local minimum value is -16

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.

We have,

Now for local minima and local maxima

At x=4,

So, x=4 is a point of local maxima.

At x=4,

At x=-4,

So, x=-4 is the point of local minimum.

At x=-4

Thus, point of local maxima is at x=4 is and its max. value is 16 and & point of local minima is at x=-4 is and its minimum value is -16.

Answer:

Point of local maxima value is and it’s local maximum value is Also point of local minima is a and it’s local minimum value is 0.

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.

Given:

Explanation:

We have,

For maxima and minima f’(x) =0

At x= a,

So, x=a is point of local minima

Now at x=a,

Also, at

Thus, point of local maxima is & its max. value is & point of local minima is a & its value is 0.

Maxima and Minima excercise 17.3 question 1 (x)

Answer:Point of local minima value is a and it’s local minimum value is 2a. Also point of local maxima is -a and it’s local maximum value is -2a

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If then is point of local minima.

If then is point of local maxima .

where & are critical points.

Put and in f(x) to get minimum value & maximum value.

Given:

Explanation:

We Have,

For maxima and minima ,f’(x)=0

Maxima and minima exercise 17.3 question 1 (xi)

Answer:

x=1 is a point of local maxima and its maximum value is 1 and x=-1 is a point of local minima and its minimum value is -1.Hint:

Putting x=1,-1in f''(x)

Given:

Explanation:

Differentiating f(x) with respect to x

These x=1 and x=-1 we the possible point of local maxima and minima

Differentiating f’(x) with respect to x

Using the formula,

So x=1 is local maxima and its maximum value f(1)=1

So , x = 1 is a point of local maxima and its minimum value f(-1)=-1

Maxima and Minima Excercise 17.3 question 1 (xii)

x= is a point of local maxima and its maximum value is 5/4.

Using chain rule of Differentiation

Differentiating f(x) with respect to x

[

Thus , is the point of local maxima and minima

Differentiating f’ with respect to x

So is a point of local maxima

The local maximum value at f(x)

is the local maxima and the local maximum value of at is .

Hint:

Using chain rule

Given:

Explanation:

Differentiating f with respect to x

Differentiating with respect to

When put x = 2 in f’’(x)

So x=2 is the local Minima

The Local Minimum value of f(x) at x =2 is 0

Put

The local maximum value of at is .

Maxima and Minima excercise 17.3 question 2 (iii)

Answer:x=1 is point of inflexion

x=-1 is the point of local minimum & is the point of local maximum

Hint:

Using chain rule of derivative

Given:

Explanation:

Differentiating f with respect to x

Thus and and are the possible points of local minima and maxima

Differentiating f’(x) with respect to x

Thus this test is fail as x=1 is point of inflexion

when

So , x = -1 is a point of local minimum

When

So, x=-1/5 is the point of local maximum

Hint:

Put?

Given:

Explanation:

Differentiating y with respect to x

Since x=1 and x=2 are the extreme values,f’(x)=0 at x=1 & 2

Then equation (1) and (2) become

Solving , (3) and (4) , we get

Putting

We need to prove that x = e is local maxima

Given:

The given function

Explanation:

Then

Put,

Differentiating with respect to

At x=e,

So x = e is a point of local maxima

Maxima and Minima exercise 17.3 question 5

Answer:x = 0 is local minima and its minimum value is 2 and x =1 is local maxima and its maximum value is -6

Given:

Explanation:

Differentiating f with respect to x

Thus x = 0 and x = -4 are the points of local maxima and minima

Differentiating f’ with respect to x

Put x = 0 in f ‘’ (x)

S,x=0is local minima. Hence,its minimum value will be f(0)=2

And x= -4 in f ‘’(x)

So x=-4 is local maxima. Hence, its maximum value will be f(-4)=-6

Maxima and Minima exercise 17.3 question 6

Answer:is a point of local minima and its local minimum value will be

is a point local maxima and its local maximum value will be

Hint:

Given:

Explanation:

Differentiating f(x) with respect to x

Thus or is possible points of local maxima and minima

Differentiating f’(x) with respect to x

So ,

is a point of Local Minima and its local maximum value will be

So is a point local maxima and its local maximum value will be

Maxima and Minima exercise 17.3 question 7

Answer:

Hint:

Put,

Given:

Explanation:

Differentiating f(x) with respect to x

Put and and as they are the extremum values hence

Answer:

Hence proved.

Hint:

Put,

Given:

Explanation:

Differentiating fx with respect to x

Differentiating f’(x) with respect to x ,

So, is a point local maxima.

Rd Sharma Class 12th exercise 17.3 is an essential part of the RD Sharma solutions for maths. The Class 12 RD Sharma chapter 17 exercise 17.3 solution will contain an important chapter of the NCERT maths book that needs to be studied well. RD Sharma Class 12 Solutions Maxima and Minima Ex 17.3 is going to be based on the chapter Maxima and minima.

Rd Sharma class 12-chapter 17 exercises 17.3 answers that are basic and simple. It will help students learn all formulae well and become skilled at the subject in general. RD Sharma class 12 solutions ex 17.3 has around 21 inquiries.

- Solutions identified with the Greatest and least upsides of a function in its space
- Definition and which means of greatest
- Definition and significance of least
- Nearby maxima
- Questions identified with Nearby minima
- Discover Solutions of First subordinate test for neighborhood maxima and minima
- Higher-request subordinate test

At Careers360, you will find all the RD Sharma solutions, so there is no need to search for extra materials elsewhere.

- These solutions are liberated from cost.
- An exceptional yet arranged exhibit of the subjects
- An itemized clarification of considerations and formulae

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

1. How to find the free pdf of RD Sharma Class 12 Solutions Maxima and Minima Ex 17.3 on the web?

Students will get the free downloadable pdf of RD Sharma Solutions for Class 12 Maths at the Career360 webpage. This site is trustworthy and will provide the updated and latest pdf.

2. Who offers the solutions to RD Sharma class 12 chapter 17 exercise 17.3?

Math experts have crafted the RD Sharma class 12 solutions chapter 17 exercise 17.3 answers. They have a clear knowledge of the subject and have created easy solutions that are accessible for students.

3. Is the RD Sharma Class 12th Exercise 17.3 Solution beneficial at all?

Indeed, the RD Sharma class 12 solutions chapter 17 exercise 17.3 book is super helpful in solving complex questions and learning modern techniques to get accurate answers.

4. How can one practice using the RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima pdf?

It's significantly suggested that class 12 RD Sharma chapter 17 exercise 17.3 solution should be used by students to check their answers and mark themselves to record their performance.

5. What are the contents of the RD Sharma Class 12 chapter 17 exercise 17.3?

There are 21 answers corresponding to the NCERT book in Class 12 RD Sharma chapter 17 exercise 17.3 solutions

Application Date:09 September,2024 - 14 November,2024

Application Date:09 September,2024 - 14 November,2024

Admit Card Date:04 October,2024 - 29 November,2024

Admit Card Date:04 October,2024 - 29 November,2024

Application Date:07 October,2024 - 05 November,2024

Get answers from students and experts

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

Accepted by more than 11,000 universities in over 150 countries worldwide

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 15th NOV'24! Trusted by 3,500+ universities globally

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

News and Notifications

Back to top